integral from 1 to 4 of 1/(x+1+sqrt(x))

Lösung

\int_{1}^{4} \frac{1}{x + 1 + x} d x

\frac{3 3 ln ( \frac{7}{3} ) - 2 ( 3 arctan ( \frac{5 3}{3} ) - π )}{3 3}

Dezimale

0.62775 \dots

Schritte zur Lösung

\int_{1}^{4} \frac{1}{x + 1 + x} d x

Wende U-Substitution an

= \int_{1}^{2} \frac{2 u}{u ^{2} + 1 + u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{2} \frac{u}{u ^{2} + 1 + u} d u

Vervollständige das Quadrat

u^{2} + 1 + u : (u + \frac{1}{2})^{2} + \frac{3}{4}

= 2 \cdot \int_{1}^{2} \frac{u}{( u + \frac{1}{2} ) ^{2} + \frac{3}{4}} d u

Wende U-Substitution an

= 2 \cdot \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 ( 2 v - 1 )}{4 v ^{2} + 3} d v

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 2 \cdot \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 v - 1}{4 v ^{2} + 3} d v

Multipliziere aus

\frac{2 v - 1}{4 v ^{2} + 3} : \frac{2 v}{4 v ^{2} + 3} - \frac{1}{4 v ^{2} + 3}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 2 (\int_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 v}{4 v ^{2} + 3} d v - \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{4 v ^{2} + 3} d v)

\int_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 v}{4 v ^{2} + 3} d v = \frac{1}{4} ln (\frac{7}{3})

\int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{4 v ^{2} + 3} d v = \frac{3 arctan ( \frac{5}{3} ) - π}{6 3}

= 2 \cdot 2 \frac{1}{4} ln (\frac{7}{3}) - \frac{3 arctan ( \frac{5}{3} ) - π}{6 3}

Vereinfache

2 \cdot 2 \frac{1}{4} ln (\frac{7}{3}) - \frac{3 arctan ( \frac{5}{3} ) - π}{6 3} : \frac{3 3 ln ( \frac{7}{3} ) - 2 ( 3 arctan ( \frac{5 3}{3} ) - π )}{3 3}

= \frac{3 3 ln ( \frac{7}{3} ) - 2 ( 3 arctan ( \frac{5 3}{3} ) - π )}{3 3}

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